Problem: Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}8x-8y &= 6 \\ 8x-2y &= -3\end{align*}$
Begin by moving the $x$ -term in the second equation to the right side of the equation. $-2y = -8x-3$ Divide both sides by $-2$ to isolate $y$ $y = {4x + \dfrac{3}{2}}$ Substitute this expression for $y$ in the first equation. $8x-8({4x + \dfrac{3}{2}}) = 6$ $8x - 32x - 12 = 6$ Simplify by combining terms, then solve for $x$ $-24x - 12 = 6$ $-24x = 18$ $x = -\dfrac{3}{4}$ Substitute $-\dfrac{3}{4}$ for $x$ back into the top equation. $8( -\dfrac{3}{4})-8y = 6$ $-6-8y = 6$ $-8y = 12$ $y = -\dfrac{3}{2}$ The solution is $\enspace x = -\dfrac{3}{4}, \enspace y = -\dfrac{3}{2}$.